Braking distance

The car travels at an average speed of 12 km/h and detects an obstacle 10 m in front of it. At 1 m in front of the obstacle it already runs 2 km/h. What is the braking distance? What is the required deceleration for stop in:

A) 1m
B) 1s?

Correct answer:

x =  12.96 m
a₂ =  0.5556 m/s²

Step-by-step explanation:

$$\begin{gathered} v_1=12\text{km/h}\to \text{m/s}=12:3.6\text{m/s}=3.33333\text{m/s} \\ {s}_1=10\text{m} \\ {s}_2=1\text{m} \\ {v}_2=2\text{km/h}\to \text{m/s}=2:3.6\text{m/s}=0.55556\text{m/s} \\ s=s_1-s_2=10-1=9\text{m} \\ s=\frac{1}{2}a{t}^2 \\ at=(v_1-v_2) \\ s=\frac{1}{2}(v_1-v_2)\cdot t \\ t=2\cdot s/(v_1-v_2)=2\cdot 9/(3.3333-0.5556)=\frac{162}{25}=6.48\text{s} \\ a=\frac{v_1-v_2}{t}=\frac{3.3333-0.5556}{6.48}=\frac{625}{1458}\doteq 0.4287{\text{m/s}}^2 \\ {t}_1=v_1/a=3.3333/0.4287=\frac{972}{125}=7.776\text{s} \\ x=\frac{1}{2}\cdot a\cdot t_1^2=\frac{1}{2}\cdot 0.4287\cdot 7.776^2=\frac{324}{25}\text{m}=12.96\text{m} \end{gathered}$$

$$\begin{gathered} t_2=2\cdot s_1/(v_1-0)=2\cdot 10/(3.3333-0)=6\text{s} \\ {a}_2=\frac{v_1}{t_2}=\frac{3.3333}{6}=\frac{5}{9}{\text{m/s}}^2=0.5556{\text{m/s}}^2 \end{gathered}$$

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